Question

How do you evaluate ${}^{3}{{P}_{1}}$?

Answer 1

Hint: We first discuss the general form of permutation and its general meaning with the help of variables. We express the mathematical notion with respect to the factorial form of ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. Then we place the values for ${}^{3}{{P}_{1}}$ as $n=3;r=1$. We complete the multiplication and find the solution.

Complete step-by-step solution:
The given mathematical expression ${}^{3}{{P}_{1}}$ is an example of permutation.
We first try to find the general form of permutation and its general meaning and then we put the values to find the solution.
The general form of permutation is ${}^{n}{{P}_{r}}$. It’s used to express the notion of choosing $r$ objects out of $n$ objects and then arranging those $r$ objects. The value of ${}^{n}{{P}_{r}}$ expresses the number of ways the permutation of those objects can be done.
The simplified form of the mathematical expression ${}^{n}{{P}_{r}}$ is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$.
Here the term $n!$ defines the notion of multiplication of first n natural numbers.
This means $n!=1\times 2\times 3\times ....\times n$.
The arrangement of those chosen objects is not considered in case of combination. That part is involved in permutation.
Now we try to find the value of ${}^{3}{{P}_{1}}$. We put the values of $n=3;r=1$ and get ${}^{3}{{P}_{1}}=\dfrac{3!}{\left( 3-1 \right)!}$.
We now solve the factorial values \[{}^{3}{{P}_{1}}=\dfrac{3!}{2!}=\dfrac{3\times 2\times 1}{2\times 1}=3\].
Therefore, the value of the combination ${}^{3}{{P}_{1}}$ is 3.

Note: There are some constraints in the form of ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}$. The general conditions are $n\ge r\ge 0;n\ne 0$. Also, we need to remember the fact that the combination happens first even though we are finding permutation. The choosing of the $r$ objects happens first, then we arrange them. The mathematical expression is ${}^{n}{{P}_{r}}=\dfrac{n!}{\left( n-r \right)!}=\dfrac{n!}{r!\times \left( n-r \right)!}\times r!={}^{n}{{C}_{r}}\times r!$.